Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $y = \dfrac{x - 7}{x + 7} \times \dfrac{x - 6}{x^2 - 13x + 42} $
First factor the quadratic. $y = \dfrac{x - 7}{x + 7} \times \dfrac{x - 6}{(x - 7)(x - 6)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ (x - 7) \times (x - 6) } { (x + 7) \times (x - 7)(x - 6) } $ $y = \dfrac{ (x - 7)(x - 6)}{ (x + 7)(x - 7)(x - 6)} $ Notice that $(x - 6)$ and $(x - 7)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ \cancel{(x - 7)}(x - 6)}{ (x + 7)\cancel{(x - 7)}(x - 6)} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $y = \dfrac{ \cancel{(x - 7)}\cancel{(x - 6)}}{ (x + 7)\cancel{(x - 7)}\cancel{(x - 6)}} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $y = \dfrac{1}{x + 7} ; \space x \neq 7 ; \space x \neq 6 $